Question

The approximate value of 

\[ \frac{1}{(2.002)^2} \]

is ______.

• A. 0.2495
• B. 0.2595
• C. 0.2095
• D. 0.2392

Hint:

Approximation formula: $f(x+h) \approx f(x) + hf'(x)$.

Answer 1

The Correct Option is A

Step 1: Define Function
Let (f(x) = \frac{1}{x^2} = x^{-2}). We need to find (f(2.002)).
Set (a = 2) and (h = 0.002).

Step 2: Find Derivative
(f'(x) = -2x^{-3} = -\frac{2}{x^3}).
(f(2) = \frac{1}{4} = 0.25).
(f'(2) = -\frac{2}{8} = -0.25).

Step 3: Use Linear Approximation
(f(a+h) \approx f(a) + h f'(a))
(f(2.002) \approx 0.25 + (0.002)(-0.25) = 0.25 - 0.0005 = 0.2495).
Final Answer: (A)